Difference between revisions of "Derivative of cosh"
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where $\cosh$ denotes the [[Cosh|hyperbolic cosine]] and $\sinh$ denotes the [[sinh|hyperbolic sine]]. | where $\cosh$ denotes the [[Cosh|hyperbolic cosine]] and $\sinh$ denotes the [[sinh|hyperbolic sine]]. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> From the definition, |
| + | $$\mathrm{cosh}(z)=\dfrac{e^z + e^{-z}}{2}$$ | ||
| + | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], and the definition of the hyperbolic sine, | ||
| + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cosh(z)=\dfrac{e^z - e^{-z}}{2}=\sinh(z),$$ | ||
| + | as was to be shown. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 19:20, 9 May 2016
Proposition: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cosh(x) = \sinh(x),$$ where $\cosh$ denotes the hyperbolic cosine and $\sinh$ denotes the hyperbolic sine.
Proof: From the definition, $$\mathrm{cosh}(z)=\dfrac{e^z + e^{-z}}{2}$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the hyperbolic sine, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \cosh(z)=\dfrac{e^z - e^{-z}}{2}=\sinh(z),$$ as was to be shown. █
